Question

http://prntscr.com/m4dlfk

Answer 1

yes

Answer 2

@nuts

Answer 3

Answer for question ii.

\(\frac{dy}{dx}=\frac{2y-x}{y-2x}\)
\(f(x,y)=\frac{dy}{dx}=\frac{2y-x}{y-2x}\)
\(f(\lambda~x,\lambda~y)=\frac{2(\lambda~y-(\lambda~x)}{\lambda~y-2(\lambda~x)}\)
\(=\frac{\lambda(2y-x)}{\lambda(y-2x)}=f(x,y)\)
Therefore,this diff. eqn is homogeneous.

\(v+x\frac{dv}{dx}=\frac{2(vx)-x}{vx-2x}=\frac{x(2v-1)}{x(v-2)}=\frac{2v-1}{v-2}\)

\(x\frac{dv}{dx}=\frac{2v-1}{v-2}-v\)

\(x\frac{dv}{dx}=\frac{2v-1-v(v-2)}{v-2}\)

\(x\frac{dv}{dx}=\frac{4v-v^2-1}{v-2}\)

\(\frac{v-2}{-v^2+4v-1}dv=\frac{1}{x}dx\)

\(\int_{}{}\frac{2-v}{v^2-4v+1}dv=\int_{}{}\frac{1}{x}dx\)

\(-\frac{In(v^2-4v+1)}{2}=In(x)+C\)

\(-In(v^2-4v+1=In~x^2+2C\)

\(v^2-4v+1=e^{(-2c-In~x^2)}\)

\(v^2-4v+1=e^{-2c}*e^{-Inx^2}\)

\(v^2-4v+1=C*x^{-2}\)

\((\frac{y}{x})^2-4(\frac{y}{x})+1=C*x^{-2}\)

\(\frac{y^2}{x^2}-4(\frac{y}{x})+1=C*x^{-2}\)

\(y^2-4xy+x^2=C\)

\(\frac{y^2}{2}-2xy+\frac{x^2}{2}=\frac{C}{2}\)

\(\frac{y^2}{2}-2xy+\frac{x^2}{2}=A\)

Answer 4

Got the same answer as 3i but not sure whether the working for \(3~ii.\) is valid.

Answer 3i

\(\frac{dM}{dy}=-2~~~,~~~\frac{dN}{dx}=-2\)
Since \(\frac{dM}{dy}=\frac{dN}{dx}\),this eqn is exact.

\(\frac{du}{dx}=x-2y=M(x,y)~~~,~~~\frac{du}{dy}=y-2x=N(x,y)\)

\(\int_{}{}du=\int_{}{}(x-2y)dx~~~,~~~\int_{}{}du=\int_{}{}(y-2x)dy\)

\(u(x,y)=\frac{x^2}{2}-2xy+\theta_1(y)~~~,~~~u(x,y)=\frac{y^2}{2}-2xy+\theta_2(x)\)

\(Compare~\theta_1(y)~~~and~~~\theta_2(x)~:~\)

\(\frac{x^2}{2}-2xy+\theta_1(y)=\frac{y^2}{2}-2xy+\theta_2(x)\)

\(\theta_1(y)=\frac{y^2}{2}\)

\(\theta_2(x)=\frac{x^2}{2}\)

\(\frac{x^2}{2}-2xy+\frac{y^2}{2}=u(x,y)\)

\(\frac{x^2}{2}-2xy+\frac{y^2}{2}=A\)

Answer 5

is it correct if let \(C=e^{-2c}\) ?

Answer 6

http://prntscr.com/m4py9w

Answer 7

\(A\) and \(C\) are constants